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n° 162 electrodynamic forces on busbars in LV systems Jean-Pierre Thierry Christophe Kilindjian Engineering graduate from the After graduating as an engineer CESI (Centre d'Etudes from the Ecole Supérieure Supérieures Industrielles) and d'Energie et des Matériaux of from the CNAM (Conservatoire Orléans in 1986, he then joined National des Arts et Métiers), Merlin Gerin in this same year he was initially employed in the as part of the Technical Section Iron and Steel industry (roll mill in the Low Voltage automation and fluid Switchboards unit. monitoring). Responsible for basic studies, Following several years he specialises in problems of dedicated to preparing and heat exchanges and developing mechanical electrodynamic withstand in LV vibration testing means, he equipment. joined Telemecanique in 1969. He then held in turn the positions of engineering and design department manager and technical Project manager for new products. He is currently in charge of developing prefabricated electric ducts. E/CT 162 first issued, october 1996 Cahier Technique Merlin Gerin n° 162 / p.2 electrodynamic forces on busbars in The importance attached to the concept of industrial dependability LV systems (safety of persons and equipment, availability of electrical power, reliability and maintenability of products) increasingly affects the design of the electrical devices used in industry (process...) and tertiary (hospitals). Their operating dependability thus contributes, often to a large context, to the dependability of the installation as a contents whole. This is the case of low voltage (LV) switchboards and of prefabricated transformer- 1. Introduction p. 4 switchboard connections. 2. Electrodynamic forces between two Preliminary remarks p. 4 This quest for dependability requires conductors: origin and calculations studies in order to master, from the Origin and calculation methods p. 4 design stage, the behaviour of their Calculation for two parallel p. 6 components in the light of their filiform conductors of infinite environment and of possible operating length stresses. One of these studies has Influence of conductor shape p. 6 already been dealt with in a «Cahier Conductors of reduced length p. 7 Technique» (thermal behaviour of LV electric switchboards). Withstand of Non-rectilinear conductors p. 7 electrodynamic forces is now the Calculation in the case of p. 7 subject of a second study. complex configurations Designers will find in this «Cahier 3. Electrodynamic forces in a Reminder on short-circuit p. 9 Technique» the calculations laid down three-phase busbar on a two or current making to allow for these forces and in three-phase fault particular to determine LV busbar Maximum force on a p. 10 requirements (prefabricated in ducts for three-phase busbar electrical power distribution, and in Resonance phenomena p. 11 switchboards). 4. Application to LV three-phase Case of busbars in p. 12 However calculation alone is not busbars LV switchboards sufficient and results need to be Case of prefabricated ducts p. 16 validated by a real-life tests. We thus of the Canalis and Victa Dis type briefly describe the standardised tests. 5. Conclusion p. 20 6. Bibliography p. 20 Cahier Technique Merlin Gerin n° 162 / p.3 1. introduction The problem of withstanding between the various conductors of a geometry of the conductors and electrodynamic forces arises on the LV installation (solid conductors of the associated structures. LV power circuits of the installation. bar, cable type...) generate considerable However a few approximations yield in Although mainly dependent on the forces (several thousands of daNm). most cases valid results on the basis of strength of the fault current, it also These forces thus need to be simple formulae. depends on the shape of the determined in order to mechanically After a brief reminder of calculation of conductors, their mutual setout and size both the conductors and the electrodynamic forces in simple securing method. Although this problem structures supporting them so that they geometries, this Cahier will deal with can be solved by calculation, only can withstand these forces whatever busbars in switchboards and validation by a real-life tests enables protective devices are placed upstream prefabricated ducts on the basis of provision of a document acknowledging and downstream (standards stipulate these formulae. conformity with standard and/or electrodynamic withstand tests of one customer requirements. second). The very high current strengths that The exact calculation of electrodynamic may occur during a short-circuit forces is often complex in view of the 2. electrodynamic forces between two conductors: origin and calculations The problem of conductor withstand to However the root mean square values whether between two current electrodynamic stresses is certainly not Irms are used in most cases; in this elements or between a magnetic field new as is shown by the number of case Irms must be multipled by a and an electric current (work publications which have treated this coefficient defined in chapter 3. conducted by Oersted, Ampère...) issue. However this problem is still of c forces are expressed in absolute resulted in the construction of a interest to designers as a result of the value without specifying their direction theoretical framework integrating application of modern numerical depending on field and current these dynamic phenomena between methods which provide a solution for direction. conductors through which electric complex conductor configurations. This In most cases they are forces per unit current flows. accounts for the summary presented in of length. The direction of the electrodynamic this chapter. c conductors are made of non- forces is known (repulsion if the magnetic material and are sufficiently currents in the conductors flow in distant from all magnetic elements preliminary remarks opposite directions, otherwise likely to alter distribution of the attraction) and their values are Application of the formulae call for magnetic field that they create. compliance with the following points: obtained by applying the laws of c skin effect and proximity phenomena c all the formulae involve the product of magnetism. which can considerably alter current the current strengths, I1.I2, flowing in There are in fact two main methods distribution in the cross-section of solid each conductor and inter-reacting. If for calculating electrodynamic forces. conductors are ignored. their values are identical, this product is The first method consists of replaced by the term I2. calculating the magnetic field created c the current strengths appearing in the origin and calculation by an electric current at a point in formulae correspond to the peak value methods space, then deducing from it the of the currents conveyed in each The highlighting and understanding a resulting force exerted on a conductor conductor. century ago of mutual influences, placed at this point and through which Cahier Technique Merlin Gerin n° 162 / p.4 an electric current flows (possibly or Ampère's theorem: The second method is based on different from the first one). → calculating the potential energy To calculate the field it uses (see box, ∫B d l = µ0 I , variation of a circuit and uses Maxwell's c fig. 1) either Biot and Savart's law: theorem : and to calculate electrodynamic force, it → → δΦ → µ dl∧ u uses Laplace's law: (4) Fx = i δx (1) d B = 0 i , → → → 4π r2 (3) d f = i d l ∧ B . (see box, figure 1). Biot and Savart's law Ampère's theorem Each element of a circuit through which a current i → Deduced from Biot and Savart's formula, it is flows, of a length d l , produces at a point M a field → expressed as follows: d B such that: Let I be the current strength flowing through a → → conductor crossing any surface of contour C. → µ0 d l ∧ u Circulation of the magnetic field along C is given by the dB = i . 4π r2 equation: This field is: → c→ perpendicular to the plane defined by the element ∫ B d l = µ0 I . c d l containing point P and point M, c oriented to the left of an observer placed on the Laplace's law element, with the current flowing from his feet to his When a circuit through which a current of strength i head and his gaze directed to point M (Ampère's → flows, is placed in a magnetic field B , each element → theorem) d l of the circuit is subjected to a force equal to: → → → → → → c modulus d B where u is the directing vector of PM . d f =i d l∧B → When B has an electric circuit as its origin, the law applied to each one expresses the force exerted between them: → → → → → i d f = i1 d l ∧ B 2 = i2 d l ∧ B1 . Maxwell's theorem → The work of the electromagnetic forces exerted during dB displacement of an undeformable conductor through → which an invariable current flows, placed in a magnetic dl → field, has the following expression: θ u w = i Φ or Φ is the flow of the magnetic field swept M P r during the displacement. Used in the form of elementary work, it easily obtains → the components Fx, Fy and Fz of the resultant F of the electromagnetic forces: dw = i dφ → → = ∫d f d l → → = F d l hence δΦ Fx = i δx and likewise for Fy and Fz. fig. 1: reminder of physical laws. Cahier Technique Merlin Gerin n° 162 / p.5 According to the geometry of the always valid. In this case the influence F/ l in N/m, conductor system considered and to of conductor shape may be determined I1 and I2 in A, the calculation difficulty, one of the by considering the conductor cross- d in m. three approaches (1)+(3), (2)+(3) or (4), section as a superimposition of inter- Examples of forces withstood by two can be used. acting current lines. This approach was parallel bars on a short-circuit are given However the results obtained may differ made by Dwight for a conductor with a in the table in figure 3. slightly according to the approach used rectangular cross-section. Although the same approach can be since the assumptions on which these The resulting corrective factor, followed for all conductor shapes, laws are established are not the same. conventionally denoted k, can be calculations quickly become tiresome. determined by calculation. However as the expression of k is relatively complex, In the above equation the term (k/d) is calculation for two parallel often replaced by 1/D, where D stands its value is determined in most cases on stranded conductors of the S-shaped curves as in figure 2. for the distance between the infinite length conductors corrected to allow for the The equation then has the form: influence of their shape. For simple geometries such as filiform F/ l = 2 10-7 I1 I2 (k/d) These coefficients are also useful in the rectilinear conductors, application of Biot and Savart's and of Laplace's law where : case of a set of three-phase conductors results in the classical formula for electrodynamic force between two current lines; F/ l = 2 10-7 I1 I2/d k where : 1.4 F/ l in N/m, a I1 and I2 in A, d in m, 1.3 b b (The coefficient 2 x 10-7 results from 0.01...0.2 a the ratio µ0/4 π). 1.2 As this formula acts as a basis d throughout this study, we must specify 1.1 the assumptions for which this 0.5 a expression is valid. b c the conductors are reduced to a 1 current line. Their cross-section is thus 1 reduced to a point. In practice this 0.9 condition is considered acceptable for conductors of all cross-sections if the 2 distance between the two conductors is 0.8 considerably larger than the largest 5 transverse dimension of the conductors 10 0.7 (e.g. ten times). 15 20 c the conductors are considered to be 30 0.6 40 rectilinear and infinitely long. In practice 50 60 this condition may be considered 80 0 10 satisfactory if they are at least 15 to 20 0.5 times longer than the distance between b a them. Whenever one of these assumptions is 0.4 not valid, a corrective factor must be applied. 0.3 influence of conductor 0.2 d a shape 2 3 4 5 6 8 10 20 40 60 80 100 200 300 This formula of F/ l only applies to current lines. However for solid Fig. 2: variation of k as a function of ratios b/a and d/a (Dwight's chart). conductors this assumption is not Cahier Technique Merlin Gerin n° 162 / p.6 containing several conductors per This formula can be used only for phase. This case is dealt with in values of a and b chapter 3. a such that: 1 < < 10 I b conductors of reduced length calculation in the case of c conductors of identical length complex configurations d When conductors have the same lengt l , 15 to 20 times smaller than their The busbar configurations considered centre distance d, the resulting force is: up to now in this study were mainly mono-dimensional, or sometimes two- l d dimensional in the case of conductors c1 l c2 d2 F = 2 10 −7 I 2 1+ − forming an angle. In these cases, the D l2 l methods used to calculate c conductors of unequal length (see electrodynamic forces result in fig. 4: drawing showing two conductors of fig. 4) relatively simple formula. unequal length. In this case the resulting force is: However conductors can be arranged in many different ways or be associated l F = 2 10 −7 I 2 D [C1+ C2] with a «disturbing» environment, such that the above formulae cease to apply. where 2 d2 c2 d2 C = 1+ + 2 − Such arrangements are referred to as c + «complex configurations». l l l 2 l2 2 c d 2 c12 d 2 C1= 1+ 1 + 2 − + l l l 2 l2 C c l and 1 ∞ 2 d 0.9 2 c d2 c22 d2 C2 = 1+ 2 + 2 − + 0.8 1 l l l2 l2 0.7 0.5 The values of C1 and C2 can be read 0.2 l 0.6 0 on the chart in figure 5. F F 0.5 If the conductors do not face each other 0.4 over the entire length, with one passing 0.3 the other, the formula applies with c1 or b I 0.2 d c2 negative. 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 l NB a If c / l = 0, the equation is F in the above paragraph. The value of the fig. 5: calculation and variation of C as a expression between square brackets is function of the ratios c / l and d / l . given directly by reading the relevant curve on the chart in figure 5. characteristics forces a b d l k I F non-rectilinear conductors This is, for example, the case of mm mm mm m kA daN/m F conductors with a bend (see fig. 6). The b 5 80 100 1 0.91 35 224 branches may inter-act with one another when a strong current passes through 5 80 100 1 0.91 80 1170 I α them. a The conductor b may pivot around point O of the fixed conductor a. Force F has O fig. 3: characteristics required to calculate the following value: the forces F between two conductors of the same length. a a b 2 1- cosα Examples of forces withstood by two parallel fig. 6: drawing showing two part of F = 2 10-7 I 2 l + 1+ 2 b b a sinα bars during a short-circuit. conductors (a and b) with a bend. Cahier Technique Merlin Gerin n° 162 / p.7 Three types of problems may then numerically solve the problems c choice of meshing parameters and arise, either separately or combined: described by differential equations. In meshing of the calculation scope with c the conductors facing one another particular the finite elements method, the type of elements chosen; at this are not all in the same plane: the initially developed for mechanical stage, the system studied is merely a problem is three-dimensional; problems, has been extended to a wide set of nodes; c the conductors are close to metal range of sectors and notably that of c definition of boundary conditions to frames which may alter the distribution electromagnetism. solve the equations; of the magnetic field surrounding them; In short, to define the calculation scope, c carrying out the calculation; c the conductors are arranged so that it this method consists of breaking down c using the results. may be necessary to allow for skin the system studied into a certain A wide range of calculation software is effect and proximity phenomena which number of elements constituted and available, differing by the categories of may considerably alter current connected with one another by points problems they can solve and the distribution in the cross-section of solid known as nodes. The quantities which reliability of the results that they yield. conductors. are of interest to us (magnetic field, For example Merlin Gerin has chosen Calculation of electrodynamic forces for stresses) are determined numerically at the ANSYS software and the three types of problems mentioned each node by salving the relevant Telemecanique Flux 2D since: above uses the general approach equations (Maxwell and elasticity). c they enable very different problems to described in the paragraph on «origin Consequently, the value of each be dealt with (thermal, mechanical, and calculation method», namely quantity studied is not known exactly at electromagnetism...), calculation first of the value and all points of the system but only at node c they are open-ended; thus their latest distribution of the magnetic field at each level. Hence the importance of ensuring versions enable different problems to point in the system, then of the stresses a good correspondence between these be paired (magnetic and mechanical or in the conductors. The problem is thus nodes and the real system, and of mechanical and thermal...). divided into two to yield a magnetic and having a sound meshing. In practice, It is true that these methods may seem a mechanical problem. calculation using this method is made cumbersome and call for considerable The basic physical laws used are up of the following stages: investment. However with thorough therefore the same. However the c choice of the analysis type mastery of the problems relating to difficulty, compared with the simple (e.g. magnetism...); modelling techniques, they allow rapid cases, lies in performing the c choice of type of elements to evaluation of the behaviour of a system calculations, as the three-dimensional describe the system; or of one of its parts other than by tests. aspect requires a numerical approach. c definition of the system geometry and This is especially appreciable during the Numerous methods have been of the calculation scope using key design and development phases when developed in recent years to points; you consider the cost of a test campaign. Cahier Technique Merlin Gerin n° 162 / p.8 3. electrodynamic forces in a three-phase busbar on a two or three-phase fault Consideration of three-phase busbar of these forces and the conductor with isolated two-phase faults which have peculiarities when designing busbars the highest mechanical stress. the advantage, in steady state, of for LV switchboards and prefabricated As the electrodynamic forces of the behaving like one or two independent ducts, and of the peculiarities relating to current are proportional to the square of single-phase networks. the establishment and type of fault, is its maximum amplitude, the short-circuit Let us consider a fault occurring on the achieved by integrating factors into the currents need to be studied. single-phase diagram in figure 8 in formule presented in chapter 2. which R and L ω are network These peculiarities are: c relative layout of phases (conductors reminder on short-circuit impedance elements. If we set as the origin of time, the moment when the in ribbon, staggered...), current making short-circuit occurs, the e.m.f. (e) of the c phase shift of currents in each phase The aim of this paragraph is to review generator has the value: with respect to one another, and specify: c type of short-circuit (two or three- c the various short-circuit types that e = 2 E sin( ω t + α ) phase), can arise in a three-phase system, c short-circuit making characteristics c the notions of symmetrical and where α is the energising angle (see (symmetrical or asymmetrical state), asymmetrical state, fig. 9) corresponding to the offset in c the peak current value, c the procedure to follow to determine time between a zero of the e.m.f. and c the alternating aspect of currents, the expression of short-circuit currents the moment when the short-circuit was hence the vibrating aspect of the and the parameters on which they made. phenomena they generate. depend. In the remainder of this section, the Ohm's law applied to the circuit yields: The short-circuit types study will consider only busbars in di ribbon, where phases 1,2, 3 are set out There are four types on a three-phase e = R i+L network. These types are shown in dt in the same plane and with the same distance between phases. figure 7. If the current is nil before the short- Expression of short-circuit currents circuit is made, the solution for this The aim is to determine, by analysing in the case of a three-phase fault equation is: the change in electrodynamic forces as a function of time and the various parameters above, the maximum value We shall now concentrate only on symmetrical three-phase faults and [ i( t ) = 2 I sin( ω t + α - ϕ ) + sin( ϕ - α ) e -t / τ ] a) L3 c) L3 R x L2 L2 L1 L1 A Zcc Z1 I" k I" k I" k I" k B fig. 8: equivalent single-phase diagram on a b) L3 d) L3 three-phase fault (see IEC 909). L2 L2 L1 L1 u u = f(t) I" k I" k ωt α a) symmetrical three-phase short-circuit. c) short-circuit between phases, with b) short-circuit between phases, isolated or earthing. two-phase. d) phase-earth short-circuit fig. 7: the various short-circuits and their currents. The direction of the arrows showing the fig. 9: representation of α known as the currents are random (see IEC 909). energising angle. Cahier Technique Merlin Gerin n° 162 / p.9 where : by side. Thus each conductor F/ l = 2 10-7 I1 I2/d Lω undergoes at a time t a force which ϕ = arctg (impedance angle) results from the algebraic addition of its the additional corrective factor which, R according to the position of the interactions with the two other L conductor under consideration, equals τ= conductors. These conductors can R have only two situations, external or 0.808 or 0.866. The maximum force is E thus generated on the central I= central: R +L2 ω 2 2 c external position, for example conductor. phase 1: c in practice, the coefficient k takes the All the factors representing the current circuit characteristics (R and L) into F1(t) = F2→1(t) + F3→1(t) variation as a function of time are then consideration: its value is between 1 grouped in the following equation: F1(t) = cF [ i1(t) i2(t) + i1(t) i3(t)/2] and 2 (see fig. 10). cF is a function of distance between [ ] Case of a two-phase short-circuit κ = sin(ω t + α - ϕ ) + sin( ϕ - α ) e-t / τ bars and bars shape. In this case i1 = - i2 and, using the c central position, for example phase 2: above formulae, we can show that the The term κ can also be calculated maximum electrodynamic forces are F2(t) = F1→2(t) - F3→2(t) using the approximate formula defined reached when α = 0 (asymmetrical by IEC 909: F2 = cF [i1(t) i2(t) - i2(t) i3(t)] state). 3R However, as seen in the above F2max ,2ph - κ = 1.02 + 0.98 e Lω paragraph, there are many cases to be considered for current expression = 2 10-7 1 ( 2 Irms,2ph κ)2 1/d The difference with the exact value is according to the value of α, ϕ, and of less than 0.6%. Remarks the type of short-circuit. The maximum force is not shown in Analysis of this function enables In actual fact, only the value of the two-phase, as it is often thought, but in definition of the symmetrical and maximum forces is required to size the three-phase. asymmetrical states of a fault busbars: this value is the highest In actual fact: (cf. «Cahier Technique» n° 158). current occurring when α = 0. In the case of a three-phase system, F2max,3ph 0.866 I 2 rms,3ph the current in each phase takes the NB: = Fa—>b = action (force) of the F2max,2ph I 2 rms,2ph form: conductor(s) of phase a on the [ however in the three-phase distribution i1( t ) = 2 Irms,3ph sin(ω t + α - ϕ ) conductor(s) of phase b. state: +sin(ϕ - α ) e-t/ τ ] Case of a three-phase short-circuit Irms,2ph = 3 I which can also be written as: The effects on the conductors take the 2 rms,3ph form: which yields the ratio: i1( t ) = 2 Irms,3ph κ F1 = 0.87 [i1(t) i2(t) + i1(t) i3(t)/2] where Irms,3ph stands for the F2 = 0.87 [i1(t) i2(t) - i2(t) i3(t)] F2max,3ph ≈ 1.15 symmetrical root mean square current The maximum force on conductors over F2max,2ph in the three phases in steady state. time is determined by the time values In view of their relative phase shift: which cancel the derivatives of these c i2 same as i1 by replacing α by expressions with respect to time: α + 2π/3 dF1/dt = 0 et dF2/dt = 0. c i3 same as i1 by replacing α by Hence, after a few calculations, where k α - 2π/3. Imax,3ph = 2 Irms,3ph κ 2.0 Finally, the electrodynamic forces thus the two equations: depend on : 1.8 c F1max ,3ph = c the initial instant of the short-circuit (via the value of α) ; 2 10- 7 0.808 ( 2 Irms,3ph κ)2 1/d 1.6 c the characteristics of the circuit (via (case of one of the conductors external 1.4 the value of ϕ) ; to the three-phase busbar) c the phase shift between the phases c F2max ,3ph = 1.2 (2π/3). 2 10-7 0.866 ( 2 Irms,3ph κ)2 1/d 1.0 R/X 0 0.2 0.4 0.6 0.8 1.0 1.2 (case of one of the conductors external maximum force on a three- to the three-phase busbar) phase busbar Note: fig. 10: variation of factor k as a function of A three-phase busbar normally c compared with the reference formula the ratio R/X. contains three conductors placed side reviewed in chapter 2 Cahier Technique Merlin Gerin n° 162 / p.10 Test organisations often demand two δy 2 δy δy 4 expression of conductor natural F( t ) = M +λ +E J 4 and three-phase tests with currents of δt 2 δt δx resonance frequencies: identical value. These test conditions 2 where: Sk E J do not correspond to real distribution ω ok = 2 M = mass of the conductor per unit of p M characteristics and result in two-phase length, where forces which are greater than three- J = moment of inertia of the cross- Sk = coefficient function of the securing phase forces. section perpendicular to the conductor methods, for example for a bar flush axis, mounted at its ends: resonance phenomena E = modulus of elasticity, Sk = (4 k - 1) π/2 ; Forces appearing on a short-circuit do λ = damping coefficient, y = distance from a point of the k = rank of resonance frequency; not form a static phenomenon, but are vibrating quantities of a frequency twice conductor with respect to its position of p = distance between the supports. that of the network or of its multiples. equilibrium or deflection, In practice, we observe that natural Conductors which have a certain x = distance from a point of the conductor frequencies, for a specific elasticity can then start to vibrate. If the conductor with respect to a fixed cross-section, depend on the vibration frequency corresponds to a bearing, longitudinal distance between supports. natural frequency for all conductors, t = time. The calculation therefore aims at resonance phenomena may occur. In where F(t) = Fo sin (2 ω t) examining whether the stress factor, this case the resulting stresses in the where: resulting from the selected distance conductors may be far greater than Fo = amplitude of the force, between the supports, is acceptable for the natural frequency of the conductor those created by the forces due to the ω = network pulsation (ω = 2 π f). peak current value. It is thus necessary or all the conductors, resulting in a The solutions take the form: coefficient R, homogenenous with a to determine the ratio between the real y = cste Fk(t) Gk(x) length: and static forces undergone by the where the functions Fk(t) and Gk(x) conductor. This ratio conventionally depend on time and on the space EJ denoted Vσ is known as the stress R=4 103 factor. In addition to the mechanical variable respectively, as well as on: M ω2 c the securing methods, The graph in figure 12 shows the stress characteristics of the conductors, we c the electrodynamic force relating to factor Vσ to be anticipated as a function must allow for the way in which they are secured in the device housing them the short-circuit state (symmetrical or of the ratio p/R, i.e. of the distance p (LV switchboard, duct...). We thus need asymmetrical). between the supports. p must be chosen to reason on the «busbar structure». The complete study was conducted by so that the ratio is outside the hatched There are two standard methods for Baltensperger and leads to an zone for the accepted factor Vσ. securing busbars: flush mounting and simple support. However in reality the insulating elements support the conductors, which results in a combination of these two methods Vσ - Vσ: stress factor (see fig. 11). The large number of parameters to be - p: distance between supports 5 considered makes a complete study of EJ - R=4 103 these phenomena complex. M ω2 The starting point for such a study is 4 the general equation applied to a conductor assumed to have an elastic behaviour: 3 resonance zone 2 a b c 1 p 1 2.24 2.45 3 3.55 4 5.22 6.12 7 R fig. 11: the various busbar securing methods: by flush mounting (a), by simple support (b) and a combination of both (c). fig. 12: stress factor Vσ to be anticipated as a function of the ratio p/R. Cahier Technique Merlin Gerin n° 162 / p.11 4. application to LV three-phase busbars In this chapter the authors define how In practice, sizing requires determination The details of each stage are described the above theoretical considerations of the distance between the supports below for a busbar consisting of several are taken into account for two LV items and thus the number required, for a rectangular cross-section bars per of equipment, namely LV switchboards specific busbar and support technology. phase. and prefabricated electrical ducts of the Canalis and Victa Dis type. Practical calculation procedure I - Basic data for carrying out the The method to be followed is calculation summarised in the chart below: c dimension and shape of a conductor case of busbars in LV (for example for a bar, its thickness a switchboards I. Definition of basic data and its width b in m.) The three-phase busbar of a LV electric c number of conductors per phase: n. switchboard is made up of a set of II. Calculation of forces conductors grouped by phase and held c root mean square value of the short- in place by supports. circuit current: Isc in kA. III. Calculation of the distance between c type of fault: two or three-phase. It is characterised by: supports based on stresses on the c the shape of the conductors, c distance between phase centres: dph conductor with the greatest stress c the relative layout of the phases, in m. c the arrangement of the conductors in c conductor securing method in the the same phase, IV. Calculation of the distance supports (flush mounting or simple c the type of support and the conductor between supports based on stresses support). securing method (insulating bars, on supports. This data is taken into account by a combs, insulating rods...). coefficient ß: The various elements making up the V. Determination of the maximum ß = ß1 for all the conductors of a busbar system must be sized to phase, distance between supports, and withstand the electrodynamic forces verification of the vibration behaviour ß = ß2 for a conductor belonging to which appear when a short-circuit of the busbar. one phase, occurs (see fig. 13). c elastic limit of the conductor: Rp0.2 in N/m2 (Rp0.2 = 125 x 106 N/m2 for 1050 type aluminium and Rp0.2 = 250 x 106 N/m2 for copper). c characteristics of supports: mechanical withstand Rm (in N/m2) according to the type of stress, and cross-section of the stressed support Sm (in m2). II - Calculation of forces Each conductor of a phase is subjected to a force due to the actions between phases and to the actions of the other conductors of the same phase. The maximum force is exerted on the most external conductors of the central phase. This conductor is subjected: c firstly to the force resulting from the other two phases: F1/ l = 0.87 (or 1) 2 10-7 k1 (2.2 Isc)2 1/dph) fig. 13: busbar of a Masterbloc LV switchboard, designed to withstand the effects of a 80 kA 0.87 : if the fault is three phase short-circuit current (Merlin Gerin). 1 : if the fault is two phase Cahier Technique Merlin Gerin n° 162 / p.12 (force per unit of length of the busbar in N/m). k1 = Dwight's coefficient allowing for the shape of all the conductors of the phase 1 phase 2 phase 3 phase. d' d ph This coefficient, parameterised by the ratios height (h)/width of a phase (I') and dph/width of a phase, can be b h calculated or read on charts. d1 → 2 Isc = root mean square value of the d1 → 3 a l' short-circuit current in kA. dph = distance between phase centres here b = h in m. k1 = f(h, I', dph) The multiplying factor 2.2 is used to k2 = f(a, b, d') calculate the peak value of the short- circuit current. c secondly to the force of attraction fig. 14: parameters considered to establish the equation for the force of attraction between (current in the same direction) resulting busbar conductors from the other conductors of the phase considered (see fig. 14), if these are mechanically linked: configuration; bars of the same phase a distance d equal to the smallest value are flush mounted and the three of d1 and d2: F2 / l = ∑ F21→i / l (in N / m) phases positioned (see fig. 15). d i min (d1,d2). i «Busbar deformation» criterion Moreover you must ensure that this Equation of the same form as the one The bar with the greatest stress must distance does not generate resonance above, but taking into account the not be deformed. However a slight phenomena. following three parameters: residual deformation is accepted This calculation procedure complies d1 → i = centre distance from according to a coefficient q defined by with the recommendations of the conductor 1 to conductor i in m, the IEC 865 standard. IEC 865 standard (1986) dealing with n = number of conductors per phase, The above formula includes d1. This calculation of the effects of short-circuit k2 = Dwight's coefficient for the phase distance between supports can be currents as regards both the thermal conductor. determined from a maximum stress and mechanical aspects. level at the conductors which must not Although these calculations do not III - Calculation of the distance be exceeded, such that σ = q.Rp0.2 (for replace real-life tests, they are vital for between supports based on stresses example q = 1.5). designing new products and for on the conductor with the greatest satisfying specific cases. IV - Calculation of the distance stress between supports based on stresses The conductor with the greatest stress on supports must withstand the stress: The supports must therefore withstand ,,,,, ,, ,,,,, σ = σ1+ σ 2 the stresses linked to the force F1. β1 (F1/ l) d12 β2 (F2 / l) d12 flush ,, ,,,,, «Support break» criterion: mounted = + conductors 8Z 8 Z0 Rm Sm d2 = F1/ l and F2/ l = forces in N/m, α F1/ l where d1 = distance between two supports α = constant whose value depends on in m, the securing method and the number of Z0 = resistance module of a bar in m3, supports. phases in Z = resistance module of a phase simple support V - Determination of the maximum in m3, distance between supports, and ß1 = 0.73 (simple support coefficient), verification of the busbar vibration ß2 = 0.5 (flush mounting coefficient). behaviour fig. 15: configuration of a busbar for These values are given by way of coefficients ß1 = 0.73 (simple support) and In order to withstand electrodynamic ß2 = 0.5 (flush mounting). guidance for a specific busbar forces, the supports must be placed at Cahier Technique Merlin Gerin n° 162 / p.13 Calculation example I. definition of basic data c conductors flat copper bars thickness a = 5 mm width b = 100 mm securing: flush mounted bars c each phase is made up of n = 3 bars, with a 5 mm spacing (d' = 10 mm) c distance between phase centres dph = 95 mm c three-phase fault Isc = 80 kA rms c elastic limit of copper Rp0.2 = 250 x 106 N/m2 c mechanical withstand of support Rm = 100 x 106 N/m2 c cross-section of support subjected to tensile stress Sm = 150 x 10-6 m2 II. calculation of forces c between phases F1/ l = 0.87 2 10-7 k1 (2.2 Isc) 2 1/dph F1/ l = k1 : Dwight's coefficient, function of the ratios b/(2n - 1) a 0.87 x 2 x 10-7x 0.873 x (2.2 x 80 x 103) 2 x 1/95 x 10-3 and dph/(2n - 1) a F1/ l = 49 530 N/m = 4 953 daN/m k1 (100/5 5, 95/5 5) = 0.873 c between bars of the same phase particularly on the external bars of the central phase F2 / l = ∑ F21,i / l F2/ l = i 2 x 10-7x (2.2 x 80 x 103/n) 2 x [0.248/10 x 10-3 + 0.419/20 x 10-3] 1 index of the first bar F2/ l = 31 490 N/m = 3 149 daN/m i = 2 and 3 index of the two other bars of the phase F2 1, i / l = 2 10-7 k21,i ( 2.2 Icc/n ) 2 1/d d1→i : distance between the axis bars 1 and i k2 1, i: Dwight's coefficient as a function of ratios b/a and d1→i /a k2 1, 2( 100/5, 10/5 ) = 0.248 k2 1, 3( 100/5, 20/5 ) = 0.419 III. Calculation of the distance between supports based on stresses on the conductor with the greatest stress (elastic limit of conductor) σ = β1 (F1/ l ) d12/8 Z + β2 (F2/ l ) d12/8 Z0 σ = 1.5 Rp0.2 d12 = 1.5 Rp0.2 / [β1 (F1/ l )/8 Z + β2 (F2/ l )/8 Z0 ] d12 = 1.5 x 250 x 106 / [ 0.5 x (49 530)/8 x 1.25 x 10-6 + β1 = β2 = 0.5 0.5 x (31 490)/8 x 4.2 x 10-7] Z0 = b a2/6 = 4.2 10-7 m3 Z = n Z0 = 3 Z0 = 1.25 10-6 m3 d1 = 0.229 m = 229 mm IV. Calculation of the distance between supports based on stresses on the supports (elastic limit of supports) d2 = Rm Sm/(F1/ l ) d2 = 100 x 106 x 150 x 10-6/(0.5 x 49 530) α = 0,5 d2 = 0.604 m = 604 mm V. determination of the maximum distance between supports d = minimum entre d1 et d2 d < 229 mm Cahier Technique Merlin Gerin n° 162 / p.14 Standards and tests be qualified by calculation from not the busbar impedance is taken into There are two test categories for LV an T.T.A. structure. account during calculation. In practice: equipment, namely: With respect to short-circuit current c calculation performed with a voltage withstand, an extrapolation method for equal to the operational voltage, at the c development tests assisting with the P.T.T.A. has been defined by the entrance to the design, technical report IEC 1117 (1992). switchboard ⇒ presumed value of Isc; c certification tests. c calculation performed, at extra-low Complete certification in short-circuit The latter are part of a set of tests current withstand requires three tests: voltage, at the end of the busbar at the known as «type tests» whose reports c a three-phase short-circuit current short-circuit point ⇒ real value. are frequently demanded for a product withstand test; It is obvious that for the same defined as a «Type tested assembly» c a withstand test for a short-circuit announced value of the short-circuit (T.T.A.). current between the neutral and the current strength, the second case is far This designation, which requires tests, nearest phase. Note that if the neutral more restrictive. The difference may thus forms an additional guarantee for has the same cross-section as the range from 20 to 30% according to the users. However, despite this constraint, other phases and if the distance circuit. manufacturers develop products which between the neutral and the nearest For the phase-neutral test, the value of allow them to valorise their know-how. phase is the same as the distance the short-circuit current corresponds to The type tests defined by the standards between phases, this test corresponds 60% of the value of the current IEC 439-1 (1992) and 2 (1987) or to a two-phase short-circuit; (prospective or real) of the three-phase NF 63-421 (1991) total 7 (439-1) and c a withstand test for a short-circuit test. 10 (439-2) respectively. between a phase and the protective Many manufacturers (including As regards short-circuit withstand, conductor. Merlin Gerin and Telemecanique) which is the subject of this document, For each test the manufacturer must currently tend to perform these tests in these standards specify both the test specify the root mean square value of real current. Moreover, to ensure that conditions to be complied with and the the short-circuit current and its duration, these tests are representative of the standardised value of the coefficient normally 1 s (to verify the thermal most unfavourable tests possible during connecting the peak value to the root constraint linked to the short-circuit a short-circuit, the following points must mean square value of the short-circuit current). be complied with: current (see fig. 16). As regards the value of the short-circuit c presence of an asymmetrical state at current for the three-phase test, two least on one of the three phases; If the system considered varies only values must be identified: the c presence of at least one joint or slightly from the reference system prospective value and the real value. fishplate on the tested busbar; (T.T.A.), it is known as a «Partially type c creation of a bolted short-circuit; tested assembly» (P.T.T.A.) and it can Their difference is due to whether or c consideration of vibrating phenomena, while maintaining the fault for at least ten cycles, i.e. 200 ms at 50 Hz; this time is often extended to 1 s to check thermal withstand at the same root mean square value of the cos ϕ n time (IEC 439-1). short-ciruit current (kA) The various test stages are: Ii 5 0.7 1.5 c calibration circuit by short-circuiting 5 < I i 10 0.5 1.7 the transformer outputs; 10 < I i 20 0.3 2 c connecting the busbar to the platform 20 < I i 50 0.25 2.1 transformer; 50 < I 0.2 2.2 c setting up the short-circuit (specific part connecting all the bars) on the busbar; fig. 16 : standardised value of the coefficient n connecting the peak value to the root mean c short test (roughly 10 ms) to square value of the short-circuit current; n correspond to coefficient 2 κ defined in chapter 3 determine busbar impedance; (ie. IEC 439-1). s 1 s withstand test on the assembly. Cahier Technique Merlin Gerin n° 162 / p.15 case of prefabricated ducts The electrodynamic forces developing conductors. Some ducts sandwich up when short-circuits occur, observe the to five conductors per phase. of the Canalis or Victa Dis laws stated above and result in the Designers can then either: type deflection of conductors between v leave the conductors of the same Construction of prefabricated three- insulators and in an overall vibration. phase grouped together, phase busbars of the Canalis or The shape and cross-section of the v or insert the elementary phase Victa Dis type, designed for current conductors result from the best conductors in an orderly manner (1-2-3) transmission and distribution (see possible balance achieved between: + (1-2-3) + (1-2-3) to obtain the fig. 17) complies with proper v temperature rise of conductors; «sandwiched» configuration (see procedures and with specific standards, v acceptable voltage drop; fig. 18). the main ones of which are v production cost. This type of design is ideal for the IEC 439-1 and 2 (international) With the following vital requirements for horizontal current distribution. and UL 857 (United States). mechanical withstand: that conductor deflection continues to be elastic (no c flattened design (1000 to 5000 A): Design In this layout, the rectangular cross- permanent deformation after a short- The techniques implemented vary section conductors coated with an circuit) and does not abnormally reduce according to the current ranges insulating sheath are kept in contact the insulation level (between phases or considered, especially for large all along the duct, just as in a cable between phases and earth) during the currents exceeding 100 A. (see fig. 18). Conductors are clamped short-circuit transient period in which There are currently three main duct the electrodynamic phenomena are to ensure the necessary heat designs: created. exchanges. v standard, In practice this is obtained by adjusting To simplify manufacture, conductors v sandwich, the distance between insulators. normally have constant thickness, and v flattened. only their width varies according to c sandwiched design (1000 to 5000 A): c standard design (100 to 800 A) Beyond a certain current, 1000 A, in nominal busbar current strength (up to The conductors are placed in a metal order to remain within acceptable heat roughly 250 mm). For high currents, envelope and maintained at regular exchange conditions and dimensions two or even three conductors per intervals by comb shaped insulators for the duct, the current of the same phase, but not sandwiched, are (see fig. 18). phase is distributed over several required. a) N L3 L2 L1 b) N L3 L2 L1 c) N L3 L2 L1 fig. 18: the various prefabricated three- phase busbar designs: standard (a), fig. 17: Canalis (Telemecanique) 3000 A electrical distribution prefabricated busbar. sandwiched (b) and flattened (c). Cahier Technique Merlin Gerin n° 162 / p.16 The electrodynamic forces (distributed various forces to which conductors are The calculations shown in the box (see loads) when a short-circuit occurs are subjected. fig. 19) evaluate the mechanical stress balanced in these busbars by the The structure studied has the following of the elementary conductors reaction of the envelope sheet metal. characteristics: (according to current direction) of Its thermal behaviour means that this c In = 3000 A, phases 1 and 2 for a phase 1/phase 2 type of design is ideally suited to c three conductors/phase, i.e. two-phase short-circuit with correction transmission of horizontal or vertical 1000 A/conductor, of the geometric incidence in current. c conductor cross-section = accordance with Dwight's chart. Distribution of electrodynamic 90 mm x 6 mm, Partial conclusions: forces c material = aluminium or copper, With the standard layout, an increase This paragraph uses a simple, concrete c distance between conductor and large dispersion of the forces example to visualise and quantify the centres = 18 mm. applied to the various conductor 1 1' 1" 2 2' 2" 3 3' 3" standard design + + + - - - d ( ) F = f i2 ,cosϕ with i = I / 3 1 ∑Fph1= F d 1 0.42 + 2 d 1 0.62 − 4 d 1 0.83 − 5 d 1 0.87 − 6 d 0.92 F = 0.19 ⇒ K1= 0.19 d ∑Fph1' = F − d 1 1 1 1 1 0.42 + 0.42 − 0.75 − 0.83 − 0.87 d 3 d 4 d 5 d F =− 0.63 ⇒ K1'= 0.63 d ∑Fph1" = F − d 1 1 1 1 1 0.42 + 0.62 − 0.62 − 0.75 − 0.83 2 d 2 d 3 d 4 d F =− 1.49 ⇒ K1"= 1.49 d 1 2 3 1' 2' 3" 1" 2" 3" sandwich design + - + - + - 1 1 ∑Fph1= F − d 1 1 1 0.42 + 0.83 − 0.87 − 1 − 1 4 d 5 d 8 d 9 d F =− 0.37 ⇒ K1= 0.37 d 1 ∑Fph2 = F + d 1 1 1 1 0.42 − 0.75 + 0.83 − 0.97 + 1 3 d 4 d 7 d 8 d F =+ 0.36 ⇒ K2 = 0.36 d fig. 19: mechanical stress of phase 1 and 2 elementary conductors. Cahier Technique Merlin Gerin n° 162 / p.17 elements are observed, whereas for the Application of this technology has the This technology is particularly used in sandwich layout, forces remain more or following practical limits: the following current ranges: less the same for each conductor v 1250 A, in branch-off, v 16 to 400 A in branch-off, element. v up to 6000 A in splicing. v 40 to 1000 A in splicing. In this example, the difference in NB : c electrodynamic force withstand mechanical stress has a ratio of 1 to 5 Some articulated bends produced in Whereas «bolted» technology in favour of the sandwich layout. the same plane use the «bolted» imposes on the elements of the Moreover, this layout offers another technology principle. structure in question the same advantage as for as voltage drop is c «contact» technology electrodynamic forces as for the concerned: «sandwiching» of phases Current conducted using parallel- busbars in LV switchboards, causes a reduction in the magnetic connected contact fingers. «contact» technology benefits from induction resultant and thus in these forces. As a first approach, the current is reactance,... i..e. in voltage drop. distributed in proportion to the number The layout normally chosen for the Branch-offs and splicing of parallel contacts. Each contact point contact fingers or «pawls» is Two technologies are normally chosen has a static force F (developed by an illustrated in the drawing in figure 20 to sample current at the branch-offs or external spring) whose sizing results on which it is clearly shown that the to conduct it in the splice bars of a from a compromise between the level currents flowing in opposing pawls prefabricated transmission and of the required contact resistance to run in the same direction. The distribution line. These are «bolted» ensure nominal current flow without electrodynamic forces (distributed technology and «contact» technology. abnormal temperature rise, and the loads) developed along the pawls friction force withstand during and calculated using the above c «bolted» technology conductor expansion. The connections are made from special methods thus tend towards an With this in mind, we should note the attraction. They consequently bolted pads provided in the equipment advantage of lubricating the elastic reinforce the contact force and design stage. contacts or of using, for mounted oppose the repulsion force of The above laws are also applied for contacts, silver/graphite type contacts which has as its origin sizing the pads and insulators. combinations. striction of the current lines in the vicinity of the contact point (see fig. 21). This is the self- compensation principle (see fig. 20). cross-section BB k is the shape factor, to be read off the chart l in figure 2, for a conductor with a global Electrodynamic force tests A cross-section a x b. Type tests, specific to ducts, are If we write Fa Fr so that compensation is defined by the IEC 439-2 and I achieved, the result is: NF C 63-411 standards. d 2 The main difference compared with u + 1 − 1 l 3 n k «LV switchboards» lies in the short- d circuit test conditions which specify A For example, for k = 0.8, the ratio I/d must that the tests must be performed on an reach: installed line no more than six metres cross-section AA B 4.6 for 1 contact jaw (n = 1) long with at least one splice joint and a 2.7 for 2 contact jaws (n = 2) bend (see sketch in fig. 22). 1.4 for 5 contact jaws (n = 5) 0.95 for 10 contact jaws (n = 10). Although it may seem interesting to increase B the number of parallel contact jaws n, we are I quickly limited by technological In the calculation, we consider the forces on considerations as well as by differences in each half-contact jaws, of cross-section resistance and reactance between adjacent R 2r a x b, grouping all the contact elements of contact jaws which do not allow even current jaws. distribution between each other such as is If n is the number of parallel contact jaws, assumed by the calculated value of I/d. A repulsion force F is exerted between the the total repulsion force on the half-contact We must therefore take a safety margin on two conductors: jaws is: the calculated value of I/d, as large as the Fr = 2 n 3 10-7 (I2n)2 number of parallel contact jaws is high. In The force of attraction which has to (In = natural log and r = contact point radius practice, there are applications of up to compensate it is: I k l 1+ d − d 2 2 2 x 12 parallel contacts which can withstand calculated with R. Holm's formula). Fa = 2 10 −7 acceptable short-term currents of the order 2 d l 2 l of 50 kA RMS - 1 s. fig. 21: example of striction of current lines in the vicinity of the contact point between two fig. 20: the self-compensation principle. cylindrical current-carrying elements Cahier Technique Merlin Gerin n° 162 / p.18 test source source connecting device sheathed busbar length maximum splice joint (at least one) imposed length =6m sheathed bend (at least one) short-circuit device fig. 22: sketch showing a prefabricated busbar line such as defined by the standards for the type tests. Cahier Technique Merlin Gerin n° 162 / p.19 5. conclusion The high electrodynamic forces It is thus advantageous for installers manufacturers can support in view of occurring on a short-circuit and the and/or users to choose equipment the necessary infrastructure and costs material damage that they can cause presenting a maximum guarantee involved. justify the importance attached to (T.T.A.) or made up of modified Design modifications from the type mechanical withstand of busbars. An standard elements, mounted in the tested cases are, however, possible.It importance all the more vital as busbar factory and tested (P.T.T.A.). is in this respect, to a certain extent, withstand failure requires at the very In both cases, the importance of testing that the calculation approach and the least replacement of these busbars and is obvious. However such tests call for manufacturer's knowhow can take over thus shutdown of the installation. considerable investment that only major from the experimental approach. 6. bibliography Standards c Operating dependability and LV c Calculation of three-phase busbar c IEC 439-1: Low-voltage switchgear electric switchboards installations in view of withstand of and controlgear assemblies. Type- Cahier Technique n° 156 - O. BOUJU electrodynamic forces tested and partially type-tested c Calculation of short-circuit currents R. MASCARIN assemblies. Cahier Technique n° 158 Revue générale de l'Electricité c IEC 439-2: Low-voltage switchgear R. CALVAS, B. DE METZ-NOBLAT, A. RGE, August 1957. and controlgear assemblies. Particular DUCLUZAUX, and G. THOMASSET c On the establishment of formulae requirements for busbar trunking designed to determine force per unit systems (busways). (NF C 63-411). Various publications length in case of a short-circuit affecting c IEC 865: Short-cicuit currents- c Elektodynamische Beanspruchung a three-phase busbar in a ribbon calculation of effects. von parallelen Leitern, arrangement. or R. MASCARIN. c IEC 909: Short-circuit current Electrodynamic effects on parallel lines Revue générale de l'Electricité calculation in three-phase a.c. systems. P. BALTENSPERGER RGE, March 1959. c IEC 909-1: Short-circuit current Bulletin Schweiz Elektotechn Verein c Uber den Einflu von calculation in three-phase a.c. systems. n° 25, 1944 Resonanzerscheingungen auf die Factors for the calculation of short- c Sectioned busbars in first and second mechanische Kurzschlu festigkeit von circuit currents in three-phase a.c. category installations biegesteifen Stromleitern. systems according to IEC 909. R. ROLS. P. SIEBER c IEC 1117: Method for assessing the Four-part paper published in the Revue AEG Mitteilungen n° 49, 1959 short-circuit withstand strength of de l'Aluminium n° 212 - 213 - 214 - 215, c Mechanical forces on current-carrying partially type-tested assemblies (PTTA). 1954 conductors. Merlin Gerin Cahiers Techniques c Transmission of high currents in LV E.D. CHARLES and MV ac current - 2nd part - Proceedings IEE, vol. 110, n° 9, c Thermal study of LV electric P. BEIGBEDER September 1963 switchboards Bulletin Etudes et Réalisations n° 43, Cahier Technique n° 145 c Electrodynamic forces appearing in 1957 electric substations on a short-circuit C. KILINDJIAN G. SCHAFFER Revue Brown Boveri, 1970. Réal.: Sodipe - Valence - Photo.: IPV - Grenoble Edition: DTE - Grenoble Cahier Technique Merlin Gerin n° 162 / p.20 10-96 - 2500 - Printing.: Clerc Printed in France

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